如何在编译时检测类型是否为shared_ptr [英] How to detect if a type is shared_ptr at compile time

问题描述

我想找到一种确定类型是否为shared_ptr的模板化方法，并以此为基础对函数进行新的专业化处理。

I want to get a templatized way of finding if a type is a shared_ptr and based on that I want to have a new specialization of a function.

示例main函数是

template <class T> inline
void CEREAL_LOAD_FUNCTION_NAME( RelaxedJSONInputArchive & ar,    NameValuePair<T> & t )
{
    std::cout << " CEREAL_LOAD_FUNCTION_NAME NameValuePair 1 " << std::endl;
     ar.setNextName( t.name );
     ar( t.value );
}

如果t.value是shared_ptr，那么我想使用其他函数。我在下面尝试过的

If t.value is shared_ptr then I want to have a different function specialization. I have tried below,

template <class T> inline
typename std::enable_if<is_pointer<T>::value, void>::type
CEREAL_LOAD_FUNCTION_NAME( RelaxedJSONInputArchive & ar, NameValuePair<T> & t )
 {
    std::cout << " CEREAL_LOAD_FUNCTION_NAME NameValuePair 2 " << std::endl;
   ar.setNextName( t.name );
   ar( t.value );
  }

但这似乎不起作用。这些是c ++ 11谷物库的一部分。我正在尝试自定义。

But it does not seem to work. These are part of c++11 cereal library. Which I am trying to customize.

推荐答案

以下内容可能会有所帮助：

the following may help:

template<typename T> struct is_shared_ptr : std::false_type {};
template<typename T> struct is_shared_ptr<std::shared_ptr<T>> : std::true_type {};

然后您可以执行以下操作以获得正确的功能：

then you can do the following to get the correct function:

template <class T> 
typename std::enable_if<is_shared_ptr<decltype(std::declval<T>().value)>::value, void>::type
func( T t )
{
    std::cout << "shared ptr" << std::endl;
}

template <class T> 
typename std::enable_if<!is_shared_ptr<decltype(std::declval<T>().value)>::value, void>::type
func( T t )
{
    std::cout << "non shared" << std::endl;
}

实时演示

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